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```					    Classification of CDMA Systems

• Synchronous: transmissions of users are
synchronized down to chip levels (difficult,
but permits orthogonal transmission).
• Asynchronous systems

Classification of CDMA Sequences
• Orthogonal Sequences
• Pseudonoise (PN) sequences
1
Desirable Properties of Sequences

• Good autocorrelation (for synchronization
and synchronous CDMA)
• Low crosscorrelation (for low multiple
access interference)
• Availability of large number of codes

2
Autocorrelation
• Degree of correspondence between a sequence and
a phase-shifted replica of itself.
N
C ( k )   an a n  k
n 1
For example let N=5, and {ak }  {1,1,1,1,1}
Then C(0)= 5
C(1)=-1-1-1+1-1=-3
C(2)=1+1+1-1-1=1
C(3)=1-1-1+1+1=1
C(4)=-1-1+1-1-1=-3
C(5)=C(0)=5
3
Cross correlation
• Degree of agreement between two different
sequences.          N
R( k )   anbn  k
n 1
E.g, let   {ak }  {1,1,1,1,1} {bk }  {1,1,1,1,1}
Then R(0)= 1+1-1-1-1=-1
R(1)=-1+1-1+1-1=-1
R(2)=-1+1+1+1+1=3
R(3)=1-1-1+1+1=1
R(4)=-1-1+1-1+1=-1
R(5)=R(0)=-1
4
Properties of Random Sequences
1. Relative frequencies of 0 and 1 are each 1/2
2. Run length like coin flipping
1/2 with length 1
1/4 with length 2
1/8 with length 3
1/2n with length n
3. If the sequence is shifted by any number of
elements, the resulting sequence will have equal
number of agreements and disagreements with the
original sequence.
5
Max Length PN Sequence Generator
ai 1ai 2 ai 3
1 0 0
0 1 0
1 0 1
1 1 0
1 1 1
ai 1 ai  2 ai  3
0 1 1
0 0 1
---------
1 0 0
0 1 0
6
Number of M Sequences
Shift reg. Length (n)   Seq. Length N  2  1   # of sequences
n

2                         3                  1
3                         7                  2
4                         15                 2
5                         31                 6
6                         63                 6
7                        127                16
8                        255                16
9                        511                48
10                       1023               60
11                       2047               176
12                       4095               144
13                       8191               630
14                      16,383              756
15                      32,767             1800
16                      65,535             2048
17                     131,071             7710
18                     262,143             8064
19                     524,287            27,594
20                    1,048,575           24,000

7
Gold Codes
M-seq. 1
1        2    3     4    5

1        2    3     4    5
M-seq. 2
Sequence 1:            1111100011011101010000100101100
Sequence 2:            1111100100110000101101010001110
0 shift combination:   0000000111101101111101110100010
1 shift combination:   0000101010111100001010000110001

30 shift combination:1000010001000101000110001101011     8
Properties of Orthogonal Sequences

• When two orthogonal sequences are
multiplied and then integrated (summed),
the result is zero. Therefore if they are used
as for channelization, theoretically there is
no MAI (except due to multipath and sync).
• Autocorrelation properties of orthogonal
sequences usually are not good.

9
0 0 
H1  [0]         H2      
0 1 
0 0 0 0 
0 1 0 1                 H N H N 
H4                  H 2N           
0 0 1 1                 H N H N 
           
0 1 1 0 
Replace zeros by -1 and observe that
the sum of products of two rows is zero.
Shifted versions may not be orthogonal.
10
Walsh Sequences of Order 16
W0   = 00000000 00000000   W8 = 01100110 01100110
W1   = 00000000 11111111   W9 = 01100110 10011001
W2   = 00001111 11110000   W10 = 01101001 10010110
W3   = 00001111 00001111   W11 = 01101001 01101001
W4   = 00111100 00111100   W12 = 01011010 01011010
W5   = 00111100 11000011   W13 = 01011010 10100101
W6   = 00110011 11001100   W14 = 01010101 10101010
W7   = 00110011 00110011   W15 = 01010101 01010101

11
Variable Length Orthogonal Codes
C8 (1)
C4 (1)  {1,1,1,1}
C2 (1)  {1,1}                             C8 (2)
C8 (3)
C4 (2)  {1,1,1,1}
C8 (4)
C1 (1)  {1}
C8 (5)
C4 (3)  {1,1,1,1}
C8 (6)
C2 (2)  {1,1}
C8 (7)
C4 (4)  {1,1,1,1}
C8 (8)
SF=1             SF=2           SF=4                       SF=8
Channel
Base PN
Walsh 1   code
User 1
data

Walsh N               S   Modulation
User N
data

13
Channel
data dependent PN code 1
Walsh
User 1
data                                Modulation

data dependent
Walsh          PN code N
User N

14
the common resource
• For handling mixed services and variable
bit rate demands, allocate power to ensure
that maximum interference is not exceeded.
• Physical channel allocation remains
unchanged although bit rate might change.
(reallocation of codes, time slots,
frequencies etc. not needed).

15
Example of power as the common resource:
multiplexing variable bit-rate users

16
DS-CDMA BER Performance
Assuming random binary sequences, for (U  1)
simultaneous users,
L N O
SNR = M
U
P  0
1

3N 2 E Q
N          b

Here, N  number of chips per bit (sequence length)
Pb  QcSNR h
With carefully selected sequences,
the performance can be better.
17
Simplified Capacity (CDMA)

In CDMA the total noise density is
Eb Rb
N o  N o  Io  N o  U
'

Ws
Eb    Eb     Eb
 ' 
No  Io No N  U Eb Rb
o
Ws

18
Processing Gain
In most practical cases the thermal noise No is
negligible in comparison to mutual interference
(also called self - noise). Then
Eb         Eb             1        Gp
                       U
No N  U Eb Rb U Rb
'                                Eb
o                             '
Ws         Ws       No
Ws Rc
     Gp is known as the processing gain
Rb Rb
Notice that here the minimum Nyquist bandwidth
is assumed for Ws (while Rappaport assumes twice
the minimum bandwidth).
19
Voice activity
If the carrier is turned off when the speaker is
silent, the self-noise reduces from
U Eb Rb/Ws
to
V U Eb Rb/Ws
where voice activity factor V is typically 0.35
to 0.4.

20
Example
(a)A CDMA system uses direct-sequence BPSK modulation
with a data rate R= 5 kbps. Assume 30 equal power data
users. Ignore the thermal noise. Determine the minimum chip
rate to obtain a BER of 10-5.
BER 10-5 requires an Eb/No of
approximately 10.

Gp           Gp
U=        30        Gp  300
Eb           10
'
No
This means a chip rate of
Rc  5  300  1.5Mcps

21
Example (ctd)
(b) Repeat the calculations if 15 users are data the
other 15 users are voice users with voice activity
factor of 0.4.
•15 voice activated users are equivalent to
0.4x15=6 users, then the total is 21 users
Gp           Gp
U=         21       G p  210
Eb           10
N o'
This means a chip rate of
Rc  5  210  105Mcps
.

22

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